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LeetCode Problem #589 Link

Problem:

Given the root of an n-ary tree, return the preorder traversal of its nodes' values.

N ary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)

Example 1
Input	root = [1,null,3,2,4,null,5,6]
Output	[1,3,5,6,2,4]
Visual

Example 2
Input	root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output	[1,2,3,6,7,11,14,4,8,12,5,9,13,10]
Visual

Constraints:

The number of nodes in the tree is in the range [0, 10^4].
0 <= Node.val <= 10^4
The height of the n-ary tree is less than or equal to 1000.

Idea/Approach:

Pre-order traversal is one type of Depth-First Search (DFS) algorithm. We can solve this problem by using the recursive DFS option. ( add the root(parent) to the result and call DFS recursively for each child)

C++ Solution:

class Solution {
public:
    vector<int> result;
    void preorderutil(Node * root) {
        result.push_back(root->val);
        int size = root->children.size();
        for(int i = 0; i< size; i++)
            preorderutil(root->children[i]);
    }
    vector<int> preorder(Node* root) {  
        if(!root)
            return {};
        preorderutil(root);
        return result;

    }
};